Introduction & Context

Thermal resistance networks are the standard way to predict steady-state heat flow through composite walls, insulation systems, and multilayer reactor linings. By treating each layer—and each surface film—as a discrete resistance, engineers can quickly estimate heat loss, locate the controlling resistance, and size heating or cooling loads. The calculation is embedded in process equipment design (reactors, storage tanks, furnaces), building envelope specifications, and refrigerated transport containers.

Methodology & Formulas

Convert all thicknesses to metres \[ z = \frac{z_{\text{mm}}}{1000} \]
Compute conduction resistance for each solid layer \[ R_{\text{cond}} = \frac{z}{k} \]
Compute convection resistance for each surface film \[ R_{\text{conv}} = \frac{1}{h} \]
Add resistances in series to obtain the overall thermal resistance per unit area \[ R_{\text{total}} = R_{\text{in}} + R_{\text{steel}} + R_{\text{foam}} + R_{\text{steel}} + R_{\text{out}} \]
Evaluate the heat flux \[ \frac{Q}{A} = \frac{\Delta T}{R_{\text{total}}} \quad \text{where} \quad \Delta T = T_{\text{in}} - T_{\text{out}} \]
Determine the temperature at any interface by subtracting the temperature drop across the upstream resistances \[ T_{\text{interface}} = T_{\text{in}} - \frac{Q}{A} \sum R_{\text{upstream}} \]

Regime	Typical \(h\) Range (W m^-2 K^-1)	Flow Criterion
Natural convection—inside buildings	5 – 15	\( \text{GrPr} \lesssim 10^9 \)
Forced convection—outside exposed walls	15 – 50	\( \text{Re} \gtrsim 10^4 \)
Wind-driven rain or forced draft	50 – 150	Site-specific

When stacking slabs of different materials, how do I calculate the overall thermal resistance for steady-state heat flow?

Treat each layer as a series resistor: R_total = Σ (thickness / thermal conductivity).
Ensure units are consistent (m, W m-1 K-1) so R ends in m² K W-1.
Add surface resistances if convection or radiation at the outer faces is significant.
Divide the overall temperature difference by R_total to obtain heat flux q = ΔT / R_total.

Which layer dominates the total resistance and how can I reduce it?

The layer with the lowest thermal conductivity or greatest thickness contributes most to R_total.
Replace that layer with a higher-conductivity material or reduce its thickness while meeting mechanical constraints.
If the dominant layer is an air gap, consider filling it with a conductive paste or foam to cut contact resistance.

How do I account for thermal contact resistance between slabs?

Measure or look up the contact resistance R_contact (m² K W-1) for the specific surface finish and pressure.
Add R_contact values at every interface to the one-dimensional series resistance stack.
Improve surface flatness, increase clamping pressure, or use thermal interface materials to lower R_contact.

Can I still use the series-resistance model when heat is generated inside one of the layers?

Yes, but split the stack into two fictitious halves at the plane of generation.
Calculate the temperature drop across each half using the appropriate fraction of the total heat flow.
Superimpose the internal generation temperature rise ΔT_gen = (q_gen · t²) / (2 k) where q_gen is volumetric heat generation, t layer thickness, k thermal conductivity.

Worked Example: Heat Loss Through an Insulated Cold-Room Wall

A food-processing plant needs to verify that a new sandwich-panel wall (thin stainless-steel liner on 150 mm polyurethane foam) will keep heat gain below 7 W m^-2 when the room is held at 22 °C and the ambient air is –18 °C. Determine the steady heat flux and the temperature at the steel–foam interface.

Knowns

Steel liner thickness: 0.8 mm
Foam insulation thickness: 150 mm
Thermal conductivity, steel: 16 W m^-1 K^-1
Thermal conductivity, foam: 0.025 W m^-1 K^-1
Inside surface coefficient: 12 W m^-2 K^-1
Outside surface coefficient: 25 W m^-2 K^-1
Inside air temperature: 22 °C
Outside air temperature: –18 °C

Step-by-Step Calculation

Convert thicknesses to metres:
z_steel = 0.8 mm = 0.0008 m
z_foam = 150 mm = 0.15 m
Compute individual thermal resistances (per unit area):
R_in = 1/h_in = 1/12 = 0.083 m² K W^-1
R_steel = z_steel/k_steel = 0.0008/16 = 5.0 × 10^-5 m² K W^-1
R_foam = z_foam/k_foam = 0.15/0.025 = 6.000 m² K W^-1
R_out = 1/h_out = 1/25 = 0.040 m² K W^-1
Sum resistances for the overall U-value:
R_total = R_in + R_steel + R_foam + R_out
R_total = 0.083 + 0.00005 + 6.000 + 0.040 = 6.123 m² K W^-1
Determine overall temperature difference:
ΔT = T_in – T_out = 22 °C – (–18 °C) = 40 K
Calculate steady heat flux:
Q/A = ΔT/R_total = 40/6.123 = 6.532 W m^-2
Find temperature at the steel–foam interface (T_i):
T_i = T_in – (Q/A)·(R_in + R_steel)
T_i = 22 – 6.532·(0.083 + 0.00005) = 21.455 °C

Final Answer

The heat gain through the wall is 6.53 W m^-2 and the steel–foam interface temperature is 21.5 °C. Both values satisfy the process requirement of staying below 7 W m^-2.

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